a-1-a-2-a-n-are-mutually-distinct-and-is-a-am-sequence-if-a-1-a-2-a-n-A-and-a-1-2-a-2-2-a-n-2-B-find-the-am-sequence-

Question Number 193199 by mnjuly1970 last updated on 07/Jun/23

a_{1}, a_{2}, \dots, a_{n} a r e m u t u a l l y d i s t i n c t

a n d i s a a m s e q u e n c e .

i f a_{1} + a_{2} + \dots + a_{n} = A

a n d

a_{1}^{2} + a_{2}^{2} + . . + a_{n}^{2} = B

f i n d t h e a m s e q u e n c e .

Answered by MM42 last updated on 07/Jun/23

l e t x_{1} = a & x_{i} - x_{i - 1} = d

\Rightarrow A = n a + \frac{n (n - 1) d}{2} \Rightarrow \frac{A^{2}}{n} = {n a}^{2} + n (n - 1) a d + \frac{n {(n - 1)}^{2} d^{2}}{4} (i)

x_{i}^{2} = {(a + (i - 1) d)}^{2} = a^{2} + 2 a d (i - 1) + {(i - 1)}^{2} d^{2}

\Rightarrow \underset{i = 1}{\sum^{n}} x_{i}^{2} = {n a}^{2} + a d n (n - 1) + \frac{n (n - 1) (2 n - 1)}{6} d^{2} (i i)

(i i) - (i) \Rightarrow \frac{n B - A^{2}}{n} = \frac{n (n^{2} - 1)}{12} d^{2}

\Rightarrow d = \pm \frac{2 \sqrt{3 (n B - A^{2})}}{n \sqrt{n^{2} - 1}} ✓

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